Question

for what value of'k' One zero of x^2 - 3x + (k+1) is double the other?​

Answer 1

Answer:

For k = 1 one zero of x² - 3x + (k + 1) is twice the other.

Explanation:

Given polynomial,

$\rm \implies \: p(x) = {x}^{2} - 3x + (k + 1)$

On comparing with the general form of a quadratic equation i.e., ax² + bx + c = 0

We get,

• a = 1
• b = -3
• c = (k + 1)

Let's say one zero as α

Then, another zero would be double of α

= 2α

$\rm \bullet \: sum \: of \: zeros = \dfrac{ - b}{a}$

$\implies \alpha + 2 \alpha = \dfrac{ - ( - 3)}{1}$

$\rm \implies 3 \alpha = 3$

$\rm \implies \alpha = \dfrac{3}{3}$

$\rm \implies \alpha = 1$

And also,

$\rm \bullet \: product \: of \: zeros = \dfrac{c}{a}$

$\rm \implies \alpha \times 2 \alpha = \dfrac{(k + 1)}{1}$

$\implies {2 \alpha }^{2} \rm = k + 1$

Substituting ‘α’ :-

$\rm \implies 2( {1)}^{2} = k + 1$

$\rm \implies 2 = k + 1$

$\rm \implies 2 - 1 = k$

$\rm \implies 1 = k$

$\rm \therefore k = 1$

_____________________

Note:

General form of a quadratic equation: ax² + bx + c

$\rm \bullet \: sum \: of \: zeros = \dfrac{ - coefficient \: of \: x}{coefficient \: of \: x^2}$

$\rm \bullet \: product \: of \: zeros = \dfrac{constant \: term }{coefficient \: of \: x^2}$

Answer 2

$\rm \gray{question}$

For what value of'k' One zero of x^2 - 3x + (k+1) is double the other?

Solution

Given polynomial.

= p(x) = x²-3x+(k+1)

= ax²+bx+c = 0

We get

= a = 1

= b = -3

= c = (k+1)

One zero = α

= 2α

Sum of zeros = -b/a

α+2α = -(-3)/1

3α = 3

α = 3/3

α = 1

product of zeros = c/a

α × 2α = (k+1)/1

2α² = k+1

Substituting 'a'

2(1)²=k+1

2=k+1

2-1=k

1=k

k=1